The floor illuminance directly below the sphere is E = π L R² / h², which is 25.1 lx for the given values. For your sliders E = 25.13 lx.
Problem statement
A small lamp having the form of a uniformly luminous sphere of radius R = 6.0 cm is suspended at a height h = 3.0 m above the floor. The luminance of the lamp is L = 2.0 × 10⁴ cd/m² and is independent of direction. Find the illuminance of the floor directly below the lamp.
We assume h is the distance from the centre of the sphere to the floor. This means the point O on the floor is h below the centre.
We assume every point of the sphere has the same luminance L in every direction. This means the surface obeys Lambert's law.
We assume only the lamp lights the floor. This means no light comes from the walls or the ceiling.
Symbols: L is the luminance of the sphere. R is its radius. h is the height of its centre above O. C is the centre of the sphere. A is a point of a ring on the sphere. α is the angle at C between the downward vertical and CA. β is the angle at O between the vertical and the line OA. OA is the distance from the ring to O.
The big idea in plain words
The sphere is not a point, so each part of its surface sends light to O at a different distance and a different slant. Cut the lower half of the sphere into rings around the vertical line through C and O. All pieces of a ring are at the same distance from O and make the same angles, so a ring gives one single dE. Add the rings from the bottom of the sphere upwards, until the rays just touch the sphere. The page cuts the visible cap into 10 rings of equal angle and gives each ring its own colour. Each ring adds its own share of E, and the coloured stack in the meter is the total. In the picture the sphere is drawn far larger than true size so that you can see the angles. The small strip at the left shows the true scale.
The solution in each frame
Lab frame
The lamp and the floor are at rest. The full solution is in the next section.
Other frames
No second frame is needed. Nothing moves.
The solution step by step
A ring at angle α from the downward vertical has radius R sin α and width R dα, so its area is
The triangle C, A, O has sides R, h and OA. By the cosine rule
The outward normal at A points along CA. The ray from A to O makes the angle α + β with this normal, and the angle β with the vertical at O. The ring sends O the flux dΦ = L dS cos(α + β) dΩ, where dΩ = dΣ cos β / OA² is the solid angle of a small floor area dΣ at O. Divide by dΣ:
From the triangle, cos(α + β) = (h cos α − R) / OA and cos β = (h − R cos α) / OA. So
Light leaves the sphere only if α + β ≤ 90°. At the limit the ray touches the sphere, the radius CA is perpendicular to AO, and cos α = R / h. So α runs from 0 to αmax = arccos(R / h).
Put y = h² + R² − 2 h R cos α, so dy = 2 h R sin α dα. The limits become (h − R)² and h² − R². After simplifying, the integrand is (1/y²)[(h² − R²)² − y²] and
Put in the numbers:
For your sliders E = 25.13 lx, with αmax = 88.854°.
A separate check adds 4000 rings of the sphere, using the true distance and both slants of each ring, and finds 25.133 lx, the same as the formula 25.13 lx.
Five methods
1. Force method
This does not suit the problem. It is about light on a surface, not about forces.
2. Energy method
Add the flux that reaches a small floor area dΣ from all the rings, and divide by dΣ. This is what the rings do.
3. Graph method
The first graph shows dE/dα for the rings. Rings near the bottom of the sphere add little because they are few, and rings near the edge add little because they are seen at a slant. The area under the curve is E. The second graph shows the running total. The third graph shows E falling as 1 / h² as the lamp is raised.
4. Calculus method
The substitution y = h² + R² − 2 h R cos α in the step-by-step section turns the long integral into a simple one.
5. Check method
A sphere of luminance L seen from far away has a cross-section π R², so as a Lambert source it has the intensity I = L π R² (that is, L times the projected area). The point-source rule gives E = I / h² = π L R² / h², the same answer as the long integral. Units: I in cd divided by h² in m² gives cd/m², which is lm/m², that is lx, because the steradian is dimensionless.
Fastest: the check method, I / h². Most reliable: the 4000-ring sum, because it needs no algebra.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Luminance L | Yes | E and I in proportion, and every ring share |
| Sphere radius R | Yes | E as R², the visible cap angle, the ring shares and the picture |
| Height h | Yes | E as 1 / h², the angles β and αmax, and the picture |
| Target illuminance (given-answer mode) | Sets the answer | Solves L |
| Probe speed | No | Only how fast the probe ring sweeps |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Luminance | L | 20000 cd/m² |
| Sphere radius | R | 6 cm |
| Height of the centre | h | 3 m |
| Ratio | R / h | 0.02 |
| Visible cap angle | αmax | 88.854° |
| Intensity of the sphere | I = L π R² | 226.2 cd |
| Floor illuminance | E | 25.13 lx |
Exam traps
- Using the area of the whole sphere, 4π R². Only the projected area π R² matters for a Lambert sphere seen from far.
- Using the surface area of a hemisphere with one slant factor. There are two slants, cos(α + β) and cos β.
- Integrating α up to 90° instead of arccos(R / h). Light cannot leave the sphere past the tangent rays.
- Measuring h to the bottom of the sphere. The height is to the centre.
- Mixing cm and m. R = 6.0 cm is 0.060 m.
- Using L as the intensity. The intensity is I = L π R² in candela.